Educational Codeforces Round 54 (Rated for Div. 2) D - Edge Deletion （最短路）

最新推荐文章于 2019-09-03 20:51:10 发布

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最新推荐文章于 2019-09-03 20:51:10 发布

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分类专栏： CodeForces 最短路

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本文详细解析了Codeforces比赛中的D题，通过两种方法解决如何在原有图中选取最多k条边，形成新图，同时保持最短路径不变的问题。介绍了使用优先队列优化的Dijkstra算法和构建树后进行BFS遍历的方法。

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题目链接：http://codeforces.com/contest/1076/problem/D

题意：给你一张图G，n个点，m条边，然后再给出一个k，表示在原先的图中最多选k条边，组成新图G1。使得G1中所节点数最多，并且1到G1各点的最短路不变。

题解：这道题两种解法：

第一种先求出最短路的路径，用这个路径再建一个树，然后从1这个点bfs跑一边，输出前k条边即可。

第二种用优先队列优化的最短路，优先队列中用来松弛的边，一定是必选边，然后依次输出就是答案。

这道题需要用堆优化的dijkstra。

ac代码：

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 3e5+5;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;

struct Edge
{
    int from, to; LL dist;       //起点,终点,距离
    Edge(int from, int to, LL dist):from(from), to(to), dist(dist) {}
};

struct Dijkstra
{
    int n, m;                 //结点数,边数(包括反向弧)
    vector<Edge> edges;       //边表。edges[e]和edges[e^1]互为反向弧
    vector<int> G[MAXN];      //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
    int vis[MAXN];            //标记数组
    LL d[MAXN];              //s到各个点的最短路
    int p[MAXN];              //上一条弧

    void init(int n)
    {
        this->n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].clear();
    }

    void AddEdge(int from, int to, int dist)
    {
        edges.emplace_back(from, to, dist);
        m = edges.size();
        G[from].push_back(m - 1);
    }

    struct HeapNode
    {
        int from; LL dist;
        bool operator < (const HeapNode& rhs) const
        {
            return rhs.dist < dist;
        }
        HeapNode(int u, LL w): from(u), dist(w) {}
    };

    void dijkstra(int s, vector<int>& ans, int k)
    {
        priority_queue<HeapNode> Q;
        for (int i = 0; i <= n; i++) d[i] = INF;
        memset(vis, 0, sizeof(vis));
        d[s] = 0;
        Q.push(HeapNode(s, 0));
        while (!Q.empty())
        {
            HeapNode x = Q.top(); Q.pop();
            int u = x.from;
            if (vis[u]) continue;
            vis[u] = true;
            ans.push_back(p[u]/2);
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge& e = edges[G[u][i]];
                if (d[e.to] > d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = G[u][i];
                    Q.push(HeapNode(e.to, d[e.to]));
                }
            }
        }
    }
}gao;

int n, m, k;
int vis[MAXN];
int main()
{
    scanf("%d%d%d", &n, &m, &k);
    memset(vis, 0, sizeof(vis));
    gao.init(n);
    while (m--)
    {
        int x, y, w; scanf("%d%d%d", &x, &y, &w);
        gao.AddEdge(x, y, w);
        gao.AddEdge(y, x, w);
    }
    vector<int> ans;
    gao.dijkstra(1, ans, k);
    k = min(k, int(ans.size())-1);
    printf("%d\n", k);
    for (int i = 1; i <= k; i++)
        printf("%d%c", ans[i]+1, i==ans.size()-1?'\n':' ');
    return 0;
}

我的没用堆优化的超时代码：

#include<bits/stdc++.h>

using namespace std;
const long long maxn = 6e5  + 10;
const long long inf = 0x3f3f3f3f;
struct Edge{
	long long u;
	long long v;
	long long w;
	long long next;
	long long id;
}edge[maxn],newedge[maxn];

struct node{
	long long pos ;
	long long val ;
};
long long dist[maxn];
long long book[maxn];
long long vis[maxn];
long long head[maxn];
long long pre[maxn];
long long newhead[maxn];
long long n,m,k,tot,newtot;
queue<long long> que;


void addedge(long long u,long long v,long long w,long long xx){
	edge[tot].u = u;
	edge[tot].v = v;
	edge[tot].w = w;
	edge[tot].id = xx;
	edge[tot].next = head[u];
	head[u] = tot++;  
}
void addnewedge(long long u,long long v,long long w,long long xx){
	newedge[newtot].u = u;
	newedge[newtot].w = w;
	newedge[newtot].v = v;
	newedge[newtot].id = xx;
	newedge[newtot].next = newhead[u];
	newhead[u] = newtot++;
}

void bfs(){
	memset(vis,0,sizeof(vis));
	queue<long long> q;
	while(!q.empty()){
		q.pop();
	}
	while(!que.empty()){
		que.pop();
	}
	q.push(1);
	vis[1] = 1;
	while(!q.empty()){
		long long now = q.front();
		q.pop();
		for(long long i = newhead[now] ; i + 1; i = newedge[i].next){
			if(vis[newedge[i].v] == 0){
				vis[newedge[i].v] = 1;
				q.push(newedge[i].v);
				que.push(newedge[i].id);
			}
		}
	}
}

void D(){
	memset(pre,-1,sizeof(book));
	priority_queue<node> Q;
	pre[1] = 1;
	for(long long i = 0 ; i <= n ; i ++){
		dist[i] = inf;
	}
	for(long long i = head[1] ; i + 1 ; i = edge[i].next ){
		dist[edge[i].v] = edge[i].w;
		pre[edge[i].v] = edge[i].id;
	}
	dist[1] = 0;
	book[1] = 1;

	for(long long i = 2 ; i <= n ; i ++){
		long long minx = inf ;
		long long minpos = 0;
		for(long long j = 1 ; j <= n ; j ++){
			if(book[j] == 0 && dist[j] < minx){
				minpos = j;
				minx = dist[j];
			}
		}
		book[minpos] = 1;
		for(long long j = head[minpos] ; j + 1 ; j = edge[j].next){
			long long u = edge[j].u;
			long long v = edge[j].v;
			long long w = edge[j].w;
			if(dist[v] > dist[u] + w){
				pre[v] = edge[j].id;
				dist[v] = dist[u] + w;
			}
		}
	}
	if(k >= n){
		printf("%lld\n",n-1);
		for(long long i = 2 ; i <= n ; i ++ )
			printf("%lld ",pre[i]);
	}
	else {
		printf("%lld\n",k);
		for(long long i = 2 ; i <= n ; i ++ ){
			long long ans = 0;
			for(long long j = 0 ; j < tot ; j ++){
				if(edge[j].id == pre[i]){
					addnewedge(edge[j].u,edge[j].v,edge[j].w,edge[j].id);
					ans++;
				//	cout << edge[j].u << " " << edge[j].v << endl;
				}
				if(ans == 2) break;
			}
		}
		bfs();
	
		while(k--){
			printf("%lld ",que.front());
			que.pop();
		}
	}
}
int main(){
	memset(head,-1,sizeof(head));
	memset(newhead,-1,sizeof(newhead));
	scanf("%lld%lld%lld",&n,&m,&k);
	for(long long i = 1 ; i <= m ; i ++){
		long long u,v,w;
		scanf("%lld%lld%lld",&u,&v,&w);
		addedge(u,v,w,i);
		addedge(v,u,w,i);
	}
	if(k == 0){
		printf("0\n");
		return 0;
	}
	D();
	return 0;
}